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3 years ago

The value of x when 2x+12/7=3x+5/4

Mathematics

2 answers:

lora16 [44]3 years ago

4 0

Answer:

Step-by-step explanation:

$2x+\frac{12}{7}=3x+\frac{5}{4}\\\\2x-3x=\frac{5}{4}-\frac{12}{7}\\\\-x=\frac{5*7}{4*7}-\frac{12*4}{7*4}\\\\-x=\frac{35}{28}-\frac{48}{28}\\\\-x=\frac{35-48}{28}\\\\-x=\frac{-13}{28}\\\\ x=\frac{13}{28}$

Murrr4er [49]3 years ago

3 0

Answer:

$x = \frac{13}{28}$

For further explanation see the picture attached...

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(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

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Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

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In point C:

Computing the directional derivative using the partial derivatives of f.

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